3.940 \(\int \frac {1}{(1+x)^{2/3} (-1+x^2)^{2/3}} \, dx\)

Optimal. Leaf size=20 \[ \frac {3 \sqrt [3]{x^2-1}}{2 (x+1)^{2/3}} \]

[Out]

3/2*(x^2-1)^(1/3)/(1+x)^(2/3)

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {651} \[ \frac {3 \sqrt [3]{x^2-1}}{2 (x+1)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + x)^(2/3)*(-1 + x^2)^(2/3)),x]

[Out]

(3*(-1 + x^2)^(1/3))/(2*(1 + x)^(2/3))

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(1+x)^{2/3} \left (-1+x^2\right )^{2/3}} \, dx &=\frac {3 \sqrt [3]{-1+x^2}}{2 (1+x)^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 23, normalized size = 1.15 \[ \frac {3 (x-1) \sqrt [3]{x+1}}{2 \left (x^2-1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 + x)^(2/3)*(-1 + x^2)^(2/3)),x]

[Out]

(3*(-1 + x)*(1 + x)^(1/3))/(2*(-1 + x^2)^(2/3))

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 14, normalized size = 0.70 \[ \frac {3 \, {\left (x^{2} - 1\right )}^{\frac {1}{3}}}{2 \, {\left (x + 1\right )}^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)^(2/3)/(x^2-1)^(2/3),x, algorithm="fricas")

[Out]

3/2*(x^2 - 1)^(1/3)/(x + 1)^(2/3)

________________________________________________________________________________________

giac [A]  time = 0.36, size = 13, normalized size = 0.65 \[ \frac {3}{2} \, {\left (-\frac {2}{x + 1} + 1\right )}^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)^(2/3)/(x^2-1)^(2/3),x, algorithm="giac")

[Out]

3/2*(-2/(x + 1) + 1)^(1/3)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 18, normalized size = 0.90 \[ \frac {3 \left (x -1\right ) \left (x +1\right )^{\frac {1}{3}}}{2 \left (x^{2}-1\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x+1)^(2/3)/(x^2-1)^(2/3),x)

[Out]

3/2*(x-1)*(x+1)^(1/3)/(x^2-1)^(2/3)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{2} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)^(2/3)/(x^2-1)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((x^2 - 1)^(2/3)*(x + 1)^(2/3)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \frac {1}{{\left (x^2-1\right )}^{2/3}\,{\left (x+1\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^2 - 1)^(2/3)*(x + 1)^(2/3)),x)

[Out]

int(1/((x^2 - 1)^(2/3)*(x + 1)^(2/3)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\left (x - 1\right ) \left (x + 1\right )\right )^{\frac {2}{3}} \left (x + 1\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+x)**(2/3)/(x**2-1)**(2/3),x)

[Out]

Integral(1/(((x - 1)*(x + 1))**(2/3)*(x + 1)**(2/3)), x)

________________________________________________________________________________________